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Paradox!

So have I. I also spent the lion's share of my professional life as a programmer. You didn't by chance work on the Patriot missile system, did you?

So .333333... is correctly called an irrational number not an infinity. Still does not change the actual math.

Steve, c'mon down man. We're all worried about you.

In the 80s I worked on Sidewindermissile sensors, SADARM, and PSTAR. PSTAR being one of the first alldigital pulsed Doppler RADARS. A portable battle field air defenseRADAR.


http://en.wikipedia.org/wiki/Sense_and_Destroy_ARMor


http://www.embeddedstar.com/press/content/2004/5/embedded14668.html


More recently about 4 years ago Iworked on a digital PID laser temperature controller for a laservideo [ppjector. It went into an ASIC. Prototyped using a MicrochipDspic. I use Microchip dsp controllers.








http://www.metacafe.com/watch/10278923/microvision_showwx_laser_pocket_projector/


I use DSP and code it. Control systems.


If you are experienced with real timecomputation like a missile there appears to be a gap between yourasserted experience and not understanding RPN does not directlyaffect accuracy. Order does. If you do enough computation you willfind order of calculation makes a deference, and it is not platformdependent it is inherent to finite digit arithmetic and digiitalcomputation.


Dou know how to multiply two floatingpoint numbers , I mean other than typing equations in a program? Howmany terms in a Taylor series do you need to get 3 decimal paceaccuracy for a sin? Is a table look up faster than a series expansionfor trig functions?


Whether you overflow by calculating 1/3or enter to overflow .3333333.. the calculator does not know thedifference. Due to overflow it rounds up when you multiply by .3 as Ishowed.


If you write a dedicated applicationyou can plan for such occurrences and handle it as you wish. Butgeneral calculators and math packs can not know before hand what itwill encounter. 1/3 or 1/5 are the same to a calculator except 1/3overflows.




BTW. The Patriot intercept performancesucked protecting Israel against Hussein's SCUDS in the first gulfwar...


Your assertion by using a calculatorapp that multiply .3333... by 10 is valid fails.
 
Jesus Christ. I leave for a day and it gets so much worse. :hitsthefan:If this is the logic underlying peoples' number concept, then it's no wonder I was having so much trouble getting them to grasp the paradox. Nearly all of mathematics is vitally dependent on precise number concepts. Disallowing completeness basically means you reject the real numbers, and that means you essentially reject all of calculus and more.

Part of me wants to create a tutorial thread on convergence and number systems, and the rest of me thinks that it won't help. :(

Math reduces to counting, addition, and subtraction with multiplication as multiple additions and division multiple subtractions.


You still have not shown how you get 42 balls in the vase or refuted the simulations I have shown.


Please do that before presuming to lecture on numerical analysis.

Oh, I wouldn't presume to lecture you. Like I said, I don't really think it would help. If you ever want to learn though, I'd be happy to teach you.

Pi is less than 4, and greater than 3. Thus it is a finite number.

What you talk about is a truncated floating point representation.


lim x -> 0 1/x is infinity or goes to infinity as a limit.
lim x -> x -> inf 1/x is zero or is approached as a limit.

In both cases 0 and inf are limits that are asymptotically approached but never reached.

lim x -> inf [5/(1 + (1/x))] is 5. We take it as 5 computationally knowing it is not exactly 5 but an asymptote.
Wrong. Using "limit" means "the asymptotical value" so lim x -> inf [5/(1 + (1/x))] is exactly 5.

A basic equation for a simple operational amplifier circuit..

In your calculator infinity is an overflow condition in a variable.
The binary representation of numbers as IEEE real numbers is not in any way a more real representation than algebraic representations.

PI can not be expressed as a finite number, it is an irrational number.

So .333333... is correctly called an irrational number not an infinity. Still does not change the actual math.

We take the asymptote to be 5.0 knowing in reality it is not exactly 5 for any finite value of x. The debate is philosophical as 5 would be used regardless.

:laughing-smiley-014

The series that results from your problem specification for balls in the vase is 9,18,27, 36,45.... and does not contain 42 .

Simple question, please show how you arrived at that number and why it conflicts with the problem statement sequence. I see invocation of set theory and Cardinal Numbers, but no actual solution.

I eagerly await your reply.
 
PI can not be expressed as a finite number, it is an irrational number.
Get your vocabulary right!
Pi is not infinite! Infinite means bigger than any number.

But you are correct thay it is an irrational number.

So .333333... is correctly called an irrational number not an infinity.
No. That is called a "rationsl" number

We take the asymptote to be 5.0 knowing in reality it is not exactly 5 for any finite value of x.
You didnt get the meaning of "limit". Did you?
 
Last edited:
So have I. I also spent the lion's share of my professional life as a programmer. You didn't by chance work on the Patriot missile system, did you?

So .333333... is correctly called an irrational number not an infinity. Still does not change the actual math.

Steve, c'mon down man. We're all worried about you.

In the 80s I worked on Sidewindermissile sensors, SADARM, and PSTAR. PSTAR being one of the first alldigital pulsed Doppler RADARS. A portable battle field air defenseRADAR.


http://en.wikipedia.org/wiki/Sense_and_Destroy_ARMor


http://www.embeddedstar.com/press/content/2004/5/embedded14668.html


More recently about 4 years ago Iworked on a digital PID laser temperature controller for a laservideo [ppjector. It went into an ASIC. Prototyped using a MicrochipDspic. I use Microchip dsp controllers.








http://www.metacafe.com/watch/10278923/microvision_showwx_laser_pocket_projector/


I use DSP and code it. Control systems.


If you are experienced with real timecomputation like a missile there appears to be a gap between yourasserted experience and not understanding RPN does not directlyaffect accuracy. Order does. If you do enough computation you willfind order of calculation makes a deference, and it is not platformdependent it is inherent to finite digit arithmetic and digiitalcomputation.


Dou know how to multiply two floatingpoint numbers , I mean other than typing equations in a program? Howmany terms in a Taylor series do you need to get 3 decimal paceaccuracy for a sin? Is a table look up faster than a series expansionfor trig functions?


Whether you overflow by calculating 1/3or enter to overflow .3333333.. the calculator does not know thedifference. Due to overflow it rounds up when you multiply by .3 as Ishowed.


If you write a dedicated applicationyou can plan for such occurrences and handle it as you wish. Butgeneral calculators and math packs can not know before hand what itwill encounter. 1/3 or 1/5 are the same to a calculator except 1/3overflows.




BTW. The Patriot intercept performancesucked protecting Israel against Hussein's SCUDS in the first gulfwar...


Your assertion by using a calculatorapp that multiply .3333... by 10 is valid fails.

You seem to be stuck on the RPN thing as if I stated that RPN matters. 'RPN Scientific' are two words which comprise the name of an application which is aware of rational numbers, and does not store the values of these numbers as floats (or doubles) and therefore is not susceptible to rounding or truncation errors, since no such operation should be happening. This is not a unique property of RPN as much as it is a property of well written code.

Back in the mid 90s I wrote my own calculator application, which incidentally happened to be RPN as well. It was completely capable of working with rational numbers, and the parser could recognize the whole part, decimal stem, and repeating decimal part of repeating decimals like 2.16... (as well as supporting fractions like 13/6). That is to say typing [2][.][1][6][.][.][.] would store the number with infinite precision. It was a tradeoff on speed vs convenience, so it wouldn't have worked well in a missile system, but it worked and worked well.

As long as no real number was introduced in the calculation there was no loss of precision regardless of the order of operations.

The limitations of the systems and toolchains you've worked with only make statements about those systems and toolchains, not about math.
 
The series that results from your problem specification for balls in the vase is 9,18,27, 36,45.... and does not contain 42 .

Simple question, please show how you arrived at that number and why it conflicts with the problem statement sequence. I see invocation of set theory and Cardinal Numbers, but no actual solution.

I eagerly await your reply.

Steve- assuming a velocity of light cutoff (can't move balls faster than the speed of light), and a .1 meter distance for placing and removing the balls, one ends up with 9*42 balls in the vase, and exactly 42 balls that have been tossed aside when removed. Assuming that one planned for the velocity of light cutoff (and it certainly seems someone planned it out), one probably would have only purchased exactly 420 balls, which might explain why there are Doritos in the vase too. Ohh, so yeah, he meant 42 balls outside of the vase, but forgot to specify which balls he was counting- the ones removed (42), or the ones in the vase (378).
 
Get your vocabulary right!
Pi is not infinite! Infinite means bigger than any number.

Hehe... I read that as he intended, not as he wrote: Pi cannot be expressed in a finite number of digits.
 
I got no response on this version which I think remove any paradoxes:

Think of the balls as being laid out along a line.

How long is the line at noon? Infinite!
Where is the line at noon? Infinitely far away from the starting point.

So what if each step we also moves all balls one step to the left? Then we obviously have infinitely many balls when noon comes.

The trick of removing balls just moves the bunch of balls away from the start. It doesnt remove the balls at all.
 
I got no response on this version which I think remove any paradoxes:

Think of the balls as being laid out along a line.

How long is the line at noon? Infinite!
Where is the line at noon? Infinitely far away from the starting point.

So what if each step we also moves all balls one step to the left? Then we obviously have infinitely many balls when noon comes.

The trick of removing balls just moves the bunch of balls away from the start. It doesnt remove the balls at all.

I answered this already. The labeling changes which answer you get - that is the paradox. Changing the labeling so the balls simply shift is not the labeling that yields 0. The labeling I gave removes each ball before noon, it doesn't shift the labels at all. There is a difference.
 
I got no response on this version which I think remove any paradoxes:

Think of the balls as being laid out along a line.

How long is the line at noon? Infinite!
Where is the line at noon? Infinitely far away from the starting point.

So what if each step we also moves all balls one step to the left? Then we obviously have infinitely many balls when noon comes.

The trick of removing balls just moves the bunch of balls away from the start. It doesnt remove the balls at all.

In this case you also end up with 0 balls. Proof: Label the positions 1, 2, 3,... and consider position n. After step n position n is empty. Therefore, in the limit, all the positions are empty. Therefore, 0 balls remain.

Alternate proof: Let A_n be the set consisting of all integers whose positions are filled at step n. So A_1 = {1, 2, 3, ...}; A_2 = {2, 3, 4, ...} and in general A_n = {n, n+1, n+ 2,...}. Notice that the sets are nested, so that A_m is contained in A_n, whenever m > n. Therefore, the limit as n goes to infinity is equivalent to the countable intersection of the A_n. That intersection is empty. Suppose that it is not empty. Then, it has a least element, n. But n is not in A_m for m > n. So n is not in the intersection, a contradiction.

You tried to just shift the balls instead of removing them, but that is a distinction that simply doesn't exist in the finite case. Consider the alternative procedure where instead of shifting all the balls, we simply remove the ball in the smallest-numbered position. In that case, the occupancy sets at each step are B_1 = {1, 2, 3,...} B_2 = {2, 3, 4,...} B_n = {n, n+1, n+2,...}, so A_n = B_n for every n. So there is no difference in the limit between removing the balls and shifting the line.

The problem with saying that the balls are infinitely far away from the starting point is that there are no integers that are infinitely far from 0 and there is no place in real space that is infinitely far from any other place. If you push an object off to infinity in finite time, it's just gone. You may object that this violates conservation of mass or something, but the conservation of mass is a theorem, and is therefore only valid when the assumptions use to prove it hold.* But the whole point here is that we're specifically violating those assumptions to achieve a strange result.

*A common way of stating a local version of a conservation law is to use a continuity equation and the divergence theorem. But the statements of the divergence theorem on Wikipedia assume that the space is compact, which is obviously not true here. I think I remember that you can extend conservation theorems to infinite Euclidean spaces and times, but where all the vector fields go to 0 as you approach infinity in space. But that is more directly violated because we're sending objects to infinity in finite time.**

**This can actually be done physically, I think. About a decade ago I attended a talk in dynamical systems, the details of which I don't remember, that there is a solution to the Newtonian 5-body problem in which all 5 bodies escape to infinity in finite time. The basic idea is that the Newtonian potential energy is 1/r, which means that for an ideal point, there is an infinite amount of potential energy available. Start with two binary systems at +- 1, and have a 5th ball orbiting the two systems. With the proper initial conditions, the 5th ball can do a sling shot around each binary system during each orbit, extracting energy from the 2 binary systems and sending all 5 balls to infinity in finite time. Now of course, this relies crucially on the parts of classical physics that are wrong in like 7 different ways, so maybe "physically" isn't the best word.
 
I got no response on this version which I think remove any paradoxes:

Think of the balls as being laid out along a line.

How long is the line at noon? Infinite!
Where is the line at noon? Infinitely far away from the starting point.

So what if each step we also moves all balls one step to the left? Then we obviously have infinitely many balls when noon comes.

The trick of removing balls just moves the bunch of balls away from the start. It doesnt remove the balls at all.

I answered this already. The labeling changes which answer you get - that is the paradox. Changing the labeling so the balls simply shift is not the labeling that yields 0. The labeling I gave removes each ball before noon, it doesn't shift the labels at all. There is a difference.

No.you havent shown that at all. Changing the labels doesnt change number of remaining balls.
 
The problem with saying that the balls are infinitely far away from the starting point is that there are no integers that are infinitely far from 0 and there is no place in real space that is infinitely far from any other place. If you push an object off to infinity in finite time, it's just gone.

No. Its just an offset. That it increases without bounds is totally insignificant. You really doesnt "push something off to infinity". You could equally say that it is the starting point that are "moved off to infinity" if you chooses to follow the balls instead of staying at the start.

And, no physics has nothing to do with this.

Can you give another , non paradoxal, example of where this "push objects away to infinity" is seen as a completely legal way of reasoning?

And the real telling thing is that this only works for a specific labeling scheme....
 
The problem with saying that the balls are infinitely far away from the starting point is that there are no integers that are infinitely far from 0 and there is no place in real space that is infinitely far from any other place. If you push an object off to infinity in finite time, it's just gone.

No. Its just an offset. That it increases without bounds is totally insignificant. You really doesnt "push something off to infinity". You could equally say that it is the starting point that are "moved off to infinity" if you chooses to follow the balls instead of staying at the start.

And, no physics has nothing to do with this.

Can you give another , non paradoxal, example of where this "push objects away to infinity" is seen as a completely legal way of reasoning?

And the real telling thing is that this only works for a specific labeling scheme....

You want an example where this sort of thing is legal in mathematics other than set theory. How about real analysis? The basic counterexample of a sequence of functions that converges pointwise but not uniformity is exactly this sort of thing. Let f_n be 0 except on the interval [n,n+1], where it's a triangle with height 1. Then it's |f_n|_infinity = 1 for all n, but it clearly converges pointwise to 0.

A more naturalistic version of that is the sequence of functions f_n(x) = 2*n*x/(x^2 + n^2). Fixing x, letting n approach infinity sends f_n(x) --> 0. But evaluating f_n(n) = 1 shows that the L_infinity norm does not approach 0 and therefore the sequence converges pointwise but not uniformly. The problem here is the peak at x = n. That peak exists for every n, but since n goes off to infinity the sequence still goes pointwise to 0.

Now, of course, there's nothing paradoxical about any of this. Indeed, despite the thread title even the topic of this thread isn't an actual paradox, it's just counter-intuitive.

The labeling scheme has nothing to do with everything. That's a convenience for us to talk about what's happening. But you can relabel the balls anyway you like. For example fix any bijection from the positive integers to themselves (you have an uncountable infinity to choose from). Then, for example, in step 1 we might add balls 2445, 90980234, 24943435956, 4, 53895743975, 748372104, 32849023804, 55, 8, 743, and remove 2445. In step 2 we add another 10 balls and remove ball 90980234, and so on. It's the same situation, but now the labeling is not helping us understand what's happening, it's just arbitrary now. And there are so many more ways to do this. The key is in the removal phase. As long as you remove every ball that's added, you will end up with 0 balls. If you want to end up with some finite number of balls, any scheme that removes all but finitely many balls will work.
 
If two real numbers are inequal there must be a third real number that lies between the two in value. What number lies between 0.999... and 1?

0.999... = 1
OK, so when does the time get to be 11:59.99....?
 
You want an example where this sort of thing is legal in mathematics other than set theory. How about real analysis? The basic counterexample of a sequence of functions that converges pointwise but not uniformity is exactly this sort of thing. Let f_n be 0 except on the interval [n,n+1], where it's a triangle with height 1. Then it's |f_n|_infinity = 1 for all n, but it clearly converges pointwise to 0.

A more naturalistic version of that is the sequence of functions f_n(x) = 2*n*x/(x^2 + n^2). Fixing x, letting n approach infinity sends f_n(x) --> 0. But evaluating f_n(n) = 1 shows that the L_infinity norm does not approach 0 and therefore the sequence converges pointwise but not uniformly. The problem here is the peak at x = n. That peak exists for every n, but since n goes off to infinity the sequence still goes pointwise to 0.

Now, of course, there's nothing paradoxical about any of this. Indeed, despite the thread title even the topic of this thread isn't an actual paradox, it's just counter-intuitive.
But that has nothing to do with the paradox of this thread: we are not interested in number of balls for a specific number (a specific x). The queastion was how many balls are left and that corresponds to an infinte integral which will not be zero in your example.
 
If two real numbers are inequal there must be a third real number that lies between the two in value. What number lies between 0.999... and 1?

0.999... = 1
OK, so when does the time get to be 11:59.99....?

As Archimedes says, at 12:00. The fact that the proposed scenario in the OP requires you to do something in an infinitesimal amount of time that in reality requires a finite amount of time does not change that fact.
 
Back to the point:
* At step 1 add numbers 1..10 and remove 1.
* At step 2 add numbers 11..20 and remove 2.
* At step 3 add numbers 21..30 and remove 3.
...

It seems to me that a set theory mathematician and a calculus mathematician will give a different answer on what happens after infinitely many steps.
* From the point of set theory, each natural number is added at some finite step, and each is removed at some later step. After infinitely many steps, the container is empty.
* From the point of calculus, at each step nine balls are added to the container, and the operation can be rephrased:
** At step 2 replace number 2 with 20, then add 11..19.
** At step 3 replace number 3 with 30, then add 21..29.
...
** Each number will be multiplied by 10 an infinite number of times; after infinitely many steps, the container has infinitely many balls, each labeled "infinity"!

----

Now on to the second puzzle (and let's treat it as a purely set theoretic problem): In each step add as many balls as the number of natural numbers (i.e. an infinite, countable set), and remove one. How to make sure that after infinitely many steps there are no (or exactly 42) balls?

In fact, this is the same proof as that ω x ω (and hence, the set of rational numbers) has the same cardinality as ω (ω = set of natural numbers).

* At each step N add pairs [N,1]; [N,2]; [N,3]; ... [N,k] for k being all natural numbers (starting from 1).
* At step 1 remove [1,1].
* At steps 2 and 3 remove [1,2] and [2,1].
* At steps 3-5 remove [1,3]; [2,2]; [3,1].
* At next four steps remove [1,4]; [2,3]; [3,2]; [4,1].
...

Oh yes, you wanted 42 balls left at the end. Easy - just in the first 42 steps add extra balls [N,0] which never get removed.

----

Now I see your  Ross–Littlewood paradox, and raise you  Thomson's lamp. There's a lamp controlled by a button; pressing it turns the lamp on if it's off, and vice versa. At 11:00 the lamp is off.
* At 11:30 I press the button and turn it on.
* At 11:45 I press the button and turn it off.
* After the next 1/8 of an hour I turn it on.
* After the next 1/16 of an hour I turn it off.
* After the next 1/32 of an hour I press the button again, etc.

At 12:01, after the button has been pressed infinitely many times, is the lamp on, or off? And does it matter if at 11:00 the lamp were on, rather than off?

----

(You could argue that supertasks - i.e. completing infinitely many steps in finite time - are inherently impossible, but Zeno would argue that even walking from point A to point B is a supertask.)


Mike Rosoft
 
Back to the point:
* At step 1 add numbers 1..10 and remove 1.
* At step 2 add numbers 11..20 and remove 2.
* At step 3 add numbers 21..30 and remove 3.
...

It seems to me that a set theory mathematician and a calculus mathematician will give a different answer on what happens after infinitely many steps.
* From the point of set theory, each natural number is added at some finite step, and each is removed at some later step. After infinitely many steps, the container is empty.
* From the point of calculus, at each step nine balls are added to the container, and the operation can be rephrased:
** At step 2 replace number 2 with 20, then add 11..19.
** At step 3 replace number 3 with 30, then add 21..29.
...
** Each number will be multiplied by 10 an infinite number of times; after infinitely many steps, the container has infinitely many balls, each labeled "infinity"!

----

Now on to the second puzzle (and let's treat it as a purely set theoretic problem): In each step add as many balls as the number of natural numbers (i.e. an infinite, countable set), and remove one. How to make sure that after infinitely many steps there are no (or exactly 42) balls?

In fact, this is the same proof as that ω x ω (and hence, the set of rational numbers) has the same cardinality as ω (ω = set of natural numbers).

* At each step N add pairs [N,1]; [N,2]; [N,3]; ... [N,k] for k being all natural numbers (starting from 1).
* At step 1 remove [1,1].
* At steps 2 and 3 remove [1,2] and [2,1].
* At steps 3-5 remove [1,3]; [2,2]; [3,1].
* At next four steps remove [1,4]; [2,3]; [3,2]; [4,1].
...

Oh yes, you wanted 42 balls left at the end. Easy - just in the first 42 steps add extra balls [N,0] which never get removed.

----

Now I see your  Ross–Littlewood paradox, and raise you  Thomson's lamp. There's a lamp controlled by a button; pressing it turns the lamp on if it's off, and vice versa. At 11:00 the lamp is off.
* At 11:30 I press the button and turn it on.
* At 11:45 I press the button and turn it off.
* After the next 1/8 of an hour I turn it on.
* After the next 1/16 of an hour I turn it off.
* After the next 1/32 of an hour I press the button again, etc.

At 12:01, after the button has been pressed infinitely many times, is the lamp on, or off? And does it matter if at 11:00 the lamp were on, rather than off?

----

(You could argue that supertasks - i.e. completing infinitely many steps in finite time - are inherently impossible, but Zeno would argue that even walking from point A to point B is a supertask.)


Mike Rosoft

The party is over you missed it....:D
 
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